# Thread: Find the sum of the convergent series?

1. ## Find the sum of the convergent series?

I figured an equation that works for this series but have no idea how to calculate it once I have the sum...none of the rules in our book seem to apply.

2. Originally Posted by CrazyLond
I figured an equation that works for this series but have no idea how to calculate it once I have the sum...none of the rules in our book seem to apply.

Try to use some algebra

$\left( \frac{1}{2}\right)^{2n-5}=\left( \frac{1}{2}\right)^{2(n-1)-3}=\left( \frac{1}{4}\right)^{n-1} \left( \frac{1}{2}\right)^{-3} =8\left( \frac{1}{4}\right)^{n-1}$

So your sum is the same as

$8\sum_{n=1}^{\infty}\left( \frac{3}{4} \right)^{n-1}$

So $r=\frac{3}{4}$

So the sum of the series $8\cdot\frac{1}{1-\frac{3}{4}}=8\cdot\frac{1}{\frac{1}{4}}=32$

P.S. I think your index starts at n=1 not at n =0.

3. Hello, CrazyLond!

$S \;=\;8 + 6 + \frac{9}{2} + \frac{27}{8} + \hdots$

Geometric series: first term $a = 8$, common ratio $r = \frac{3}{4}$

Its sum is: . $S \;=\;\frac{8}{1 - \frac{3}{4}} \;=\;32$

4. Originally Posted by CrazyLond
I figured an equation that works for this series but have no idea how to calculate it once I have the sum...none of the rules in our book seem to apply.

I find it easier to look at these like this:

$a_0 = 8$

$a_n = \frac 34a_{n-1}$

Looking at that a little bit (you can do the first few a_n if it helps to see, then you will be familiar enough to spot it right off) it should be clear that $a_n = 8*\left(\frac 34\right)^n$

And since you want to sum all of them, you get

$\sum_{n=0}^\infty a_n$

$=\sum_{n=0}^\infty 8*\left(\frac 34\right)^n$

$=8\sum_{n=0}^\infty \left(\frac 34\right)^n$

At which point you have the same geometric series that TheEmptySet and Soroban solved for.